# LTS Termination Proof

by AProVE

## Input

Integer Transition System
• Initial Location: l5, l4, l7, l6, l1, l3, l0
• Transitions: (pre-variables and post-variables)  l0 1 l1: x1 = ___disjvr_0HAT0 ∧ x2 = _nd_12HAT0 ∧ x3 = _rt_11HAT0 ∧ x4 = _rv_15HAT0 ∧ x5 = _st_14HAT0 ∧ x6 = _st_16HAT0 ∧ x7 = _x_13HAT0 ∧ x8 = _y_17HAT0 ∧ x1 = ___disjvr_0HATpost ∧ x2 = _nd_12HATpost ∧ x3 = _rt_11HATpost ∧ x4 = _rv_15HATpost ∧ x5 = _st_14HATpost ∧ x6 = _st_16HATpost ∧ x7 = _x_13HATpost ∧ x8 = _y_17HATpost ∧ _y_17HAT0 = _y_17HATpost ∧ _x_13HAT0 = _x_13HATpost ∧ _st_16HAT0 = _st_16HATpost ∧ _st_14HAT0 = _st_14HATpost ∧ _rv_15HAT0 = _rv_15HATpost ∧ _rt_11HAT0 = _rt_11HATpost ∧ _nd_12HAT0 = _nd_12HATpost ∧ ___disjvr_0HAT0 = ___disjvr_0HATpost l1 2 l2: x1 = _x ∧ x2 = _x1 ∧ x3 = _x2 ∧ x4 = _x3 ∧ x5 = _x4 ∧ x6 = _x5 ∧ x7 = _x6 ∧ x8 = _x7 ∧ x1 = _x8 ∧ x2 = _x9 ∧ x3 = _x10 ∧ x4 = _x11 ∧ x5 = _x12 ∧ x6 = _x13 ∧ x7 = _x14 ∧ x8 = _x15 ∧ _x7 = _x15 ∧ _x6 = _x14 ∧ _x5 = _x13 ∧ _x4 = _x12 ∧ _x3 = _x11 ∧ _x1 = _x9 ∧ _x = _x8 ∧ _x10 = _x4 ∧ _x6 ≤ 0 l1 3 l3: x1 = _x16 ∧ x2 = _x17 ∧ x3 = _x18 ∧ x4 = _x19 ∧ x5 = _x20 ∧ x6 = _x21 ∧ x7 = _x22 ∧ x8 = _x23 ∧ x1 = _x24 ∧ x2 = _x25 ∧ x3 = _x26 ∧ x4 = _x27 ∧ x5 = _x28 ∧ x6 = _x29 ∧ x7 = _x30 ∧ x8 = _x31 ∧ 1 ≤ _x22 ∧ _x32 = _x32 ∧ _x27 = _x32 ∧ _x25 = _x25 ∧ 0 ≤ _x27 ∧ _x27 ≤ 0 ∧ _x31 = −1 + _x23 ∧ _x29 = _x29 ∧ 2 ≤ _x31 ∧ _x16 = _x24 ∧ _x18 = _x26 ∧ _x20 = _x28 ∧ _x22 = _x30 l3 4 l1: x1 = _x33 ∧ x2 = _x34 ∧ x3 = _x35 ∧ x4 = _x36 ∧ x5 = _x37 ∧ x6 = _x38 ∧ x7 = _x39 ∧ x8 = _x40 ∧ x1 = _x41 ∧ x2 = _x42 ∧ x3 = _x43 ∧ x4 = _x44 ∧ x5 = _x45 ∧ x6 = _x46 ∧ x7 = _x47 ∧ x8 = _x48 ∧ _x40 = _x48 ∧ _x39 = _x47 ∧ _x38 = _x46 ∧ _x37 = _x45 ∧ _x36 = _x44 ∧ _x35 = _x43 ∧ _x34 = _x42 ∧ _x33 = _x41 l1 5 l5: x1 = _x49 ∧ x2 = _x50 ∧ x3 = _x51 ∧ x4 = _x52 ∧ x5 = _x53 ∧ x6 = _x54 ∧ x7 = _x55 ∧ x8 = _x56 ∧ x1 = _x57 ∧ x2 = _x58 ∧ x3 = _x59 ∧ x4 = _x60 ∧ x5 = _x61 ∧ x6 = _x62 ∧ x7 = _x63 ∧ x8 = _x64 ∧ 1 ≤ _x55 ∧ _x65 = _x65 ∧ _x60 = _x65 ∧ _x58 = _x58 ∧ _x49 = _x57 ∧ _x51 = _x59 ∧ _x53 = _x61 ∧ _x54 = _x62 ∧ _x55 = _x63 ∧ _x56 = _x64 l5 6 l6: x1 = _x66 ∧ x2 = _x67 ∧ x3 = _x68 ∧ x4 = _x69 ∧ x5 = _x70 ∧ x6 = _x71 ∧ x7 = _x72 ∧ x8 = _x73 ∧ x1 = _x74 ∧ x2 = _x75 ∧ x3 = _x76 ∧ x4 = _x77 ∧ x5 = _x78 ∧ x6 = _x79 ∧ x7 = _x80 ∧ x8 = _x81 ∧ _x73 = _x81 ∧ _x72 = _x80 ∧ _x71 = _x79 ∧ _x70 = _x78 ∧ _x69 = _x77 ∧ _x68 = _x76 ∧ _x67 = _x75 ∧ _x66 = _x74 ∧ _x74 = _x66 l6 7 l4: x1 = _x82 ∧ x2 = _x83 ∧ x3 = _x84 ∧ x4 = _x85 ∧ x5 = _x86 ∧ x6 = _x87 ∧ x7 = _x88 ∧ x8 = _x89 ∧ x1 = _x90 ∧ x2 = _x91 ∧ x3 = _x92 ∧ x4 = _x93 ∧ x5 = _x94 ∧ x6 = _x95 ∧ x7 = _x96 ∧ x8 = _x97 ∧ _x96 = −1 + _x88 ∧ _x98 = _x98 ∧ _x97 = _x98 ∧ _x91 = _x91 ∧ _x82 = _x90 ∧ _x84 = _x92 ∧ _x85 = _x93 ∧ _x86 = _x94 ∧ _x87 = _x95 l4 8 l1: x1 = _x99 ∧ x2 = _x100 ∧ x3 = _x101 ∧ x4 = _x102 ∧ x5 = _x103 ∧ x6 = _x104 ∧ x7 = _x105 ∧ x8 = _x106 ∧ x1 = _x107 ∧ x2 = _x108 ∧ x3 = _x109 ∧ x4 = _x110 ∧ x5 = _x111 ∧ x6 = _x112 ∧ x7 = _x113 ∧ x8 = _x114 ∧ _x106 = _x114 ∧ _x105 = _x113 ∧ _x104 = _x112 ∧ _x103 = _x111 ∧ _x102 = _x110 ∧ _x101 = _x109 ∧ _x100 = _x108 ∧ _x99 = _x107 l7 9 l0: x1 = _x115 ∧ x2 = _x116 ∧ x3 = _x117 ∧ x4 = _x118 ∧ x5 = _x119 ∧ x6 = _x120 ∧ x7 = _x121 ∧ x8 = _x122 ∧ x1 = _x123 ∧ x2 = _x124 ∧ x3 = _x125 ∧ x4 = _x126 ∧ x5 = _x127 ∧ x6 = _x128 ∧ x7 = _x129 ∧ x8 = _x130 ∧ _x122 = _x130 ∧ _x121 = _x129 ∧ _x120 = _x128 ∧ _x119 = _x127 ∧ _x118 = _x126 ∧ _x117 = _x125 ∧ _x116 = _x124 ∧ _x115 = _x123

## Proof

### 1 Switch to Cooperation Termination Proof

We consider the following cutpoint-transitions:
 l5 l5 l5: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 l4 l4 l4: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 l7 l7 l7: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 l6 l6 l6: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 l1 l1 l1: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 l3 l3 l3: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8 l0 l0 l0: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8
and for every transition t, a duplicate t is considered.

### 2 SCC Decomposition

We consider subproblems for each of the 1 SCC(s) of the program graph.

### 2.1 SCC Subproblem 1/1

Here we consider the SCC { l5, l4, l6, l1, l3 }.

### 2.1.1 Transition Removal

We remove transition 5 using the following ranking functions, which are bounded by 0.

 l1: −1 + x7 l3: −1 + x7 l4: −1 + x7 l6: −2 + x7 l5: −2 + x7

### 2.1.2 Transition Removal

We remove transitions 8, 7, 6 using the following ranking functions, which are bounded by 0.

 l1: −1 l3: −1 l4: 0 l6: 1 l5: 2

### 2.1.3 Transition Removal

We remove transition 3 using the following ranking functions, which are bounded by 0.

 l1: −1 + x7 + x8 l3: −1 + x7 + x8

### 2.1.4 Transition Removal

We remove transition 4 using the following ranking functions, which are bounded by 0.

 l3: 0 l1: −1

### 2.1.5 Trivial Cooperation Program

There are no more "sharp" transitions in the cooperation program. Hence the cooperation termination is proved.

## Tool configuration

AProVE

• version: AProVE Commit ID: unknown
• strategy: Statistics for single proof: 100.00 % (8 real / 0 unknown / 0 assumptions / 8 total proof steps)